00:01
Let's determine the absolute maximum and absolute minimum of this function f of x in the interval negative 2 .3.
00:09
And so first we have to determine the critical point of the function so that later on we can determine the values of function at the end points as well as the critical point.
00:18
And then decide the absolute maximum and absolute minimum of this function.
00:23
So let's start to find the critical point of this function.
00:26
And for that, we should set up the derivative function equals 0.
00:31
That is, we should set up f prime of x equals 0 and solve for x.
00:36
Let's find the derivative of f of x, which is represented as f prime of x.
00:42
And this equals the derivative of 2x cube is 6x squared, minus the derivative of negative 3x squared is negative 6x.
00:52
And then the derivative of negative 12 x is negative 12.
00:56
And we should set this up equals 0.
00:58
So let's set this up equal 0.
01:00
This means we can simplify this equation by dividing both size by 6.
01:06
So this is reduced to x squared minus x minus 2 equals 0.
01:12
And this quadratic equation is factorable.
01:15
This factors nicely into x minus 2 times x plus 1.
01:21
This equals 0.
01:23
So we see that from this we get x equals 0.
01:28
That is if we set this up equals 0, we get x equal to 2.
01:32
And when we set this up expression equals 0, we get x equals negative 1.
01:38
So these are the two critical points of the function f of x.
01:42
Let's see that these two lies in the interval.
01:45
Two lies in negative 2 comma 3 and negative 1 also lies in this interval.
01:51
So we are going to create a table of values for the end points as plus the critical points...